If \(\tan(A+B) = \sqrt{3}\) and \(\tan(A-B) = \frac{1}{\sqrt{3}}\), find A and B.
Answer: A
From the given information:
1) \(\tan(A+B) = \sqrt{3} \Rightarrow A+B = 60°\)
2) \(\tan(A-B) = \frac{1}{\sqrt{3}} \Rightarrow A-B = 30°\)
We have a system of two linear equations. Adding them gives:
2A = 90° => A = 45°.
Substituting A=45° into the first equation: 45° + B = 60° => B = 15°.
Please log in or register to report this question.
The value of \(2\tan^2 45° + \cos^2 30° - \sin^2 60°\) is:
Answer: B
We substitute the standard trigonometric values:
\(\tan 45° = 1\)
\(\cos 30° = \frac{\sqrt{3}}{2}\)
\(\sin 60° = \frac{\sqrt{3}}{2}\)
The expression is \(2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2 = 2(1) + \frac{3}{4} - \frac{3}{4} = 2\).
Please log in or register to report this question.
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is:
Answer: C
Be careful with the angle. The angle is with the *wall*, not the ground.
Let h be the height of the wall (adjacent side to the 60° angle) and L be the length of the ladder (hypotenuse).
\(\cos(\text{angle}) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
\(\cos 60° = \frac{h}{15}\)
\(\frac{1}{2} = \frac{h}{15}\)
h = 15 / 2 = 7.5 meters.
Please log in or register to report this question.
If \(\sin(A - B) = 1/2\) and \(\cos(A + B) = 1/2\), the value of B is:
Answer: A
From the given information:
1) \(\sin(A - B) = 1/2 \Rightarrow A - B = 30°\)
2) \(\cos(A + B) = 1/2 \Rightarrow A + B = 60°\)
We have two equations. To find B, subtract the first equation from the second:
(A + B) - (A - B) = 60° - 30°
2B = 30°
B = 15°.
Please log in or register to report this question.
The value of \(\frac{1-\tan^2 45°}{1+\tan^2 45°}\) is:
Answer: D
We know that \(\tan 45° = 1\).
Substitute this value into the expression:
\(\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\).
Alternatively, this is the formula for \(\cos(2\theta)\), so it equals \(\cos(2 \times 45°) = \cos 90° = 0\).
Please log in or register to report this question.
The top of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with the horizontal, then the length of the wire is:
Answer: D
Imagine a right-angled triangle formed by the wire (hypotenuse), the horizontal distance, and the difference in the heights of the poles (opposite side).
The difference in height = 20 m - 14 m = 6 m.
The angle of the wire with the horizontal is 30°.
\(\sin 30° = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{Height Difference}}{\text{Wire Length}}\)
\(\frac{1}{2} = \frac{6}{\text{Wire Length}}\)
Wire Length = 6 × 2 = 12 meters.
Please log in or register to report this question.
What is the minimum value of \(\sin^2\theta + \cos^2\theta\)?
Answer: B
This is a trick question based on the fundamental Pythagorean identity.
For any angle \(\theta\), the expression \(\sin^2\theta + \cos^2\theta\) is always equal to 1.
Therefore, its minimum value, maximum value, and average value are all 1.
Please log in or register to report this question.
If a triangle ABC is right angled at C, then the value of \(\cos(A+B)\) is:
Answer: A
In a triangle the angles add to 180°, and \( C = 90^{\circ} \), so
\[ A + B = 90^{\circ} \]
\[ \cos(A+B) = \cos 90^{\circ} = 0 \]
Please log in or register to report this question.
The value of \(\sin(2A)\) is given by:
Answer: C
The double-angle identity for sine is
\[ \sin 2A = 2\sin A \cos A \]
The option \( \cos^{2}A - \sin^{2}A \) is \( \cos 2A \), the companion identity - the pair most easily transposed.
Please log in or register to report this question.
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 40 m from the bank, he finds the angle to be 30°. The height of the tree is:
Answer: B
Let h be the height of the tree and x be the initial distance from the tree.
From the first observation: \(\tan 60° = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}\).
From the second observation: \(\tan 30° = \frac{h}{x+40} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x+40} \Rightarrow x+40 = h\sqrt{3}\).
Substitute h from the first equation into the second: \(x+40 = (x\sqrt{3})\sqrt{3} = 3x\).
\(40 = 2x \Rightarrow x = 20\) m.
Now find the height: \(h = x\sqrt{3} = 20\sqrt{3}\) m.
Please log in or register to report this question.
